Suppose that \(x\) is a smooth real-valued function of the real number \(t\), satisfying \(dx/dt\le b(t)x(t)\) for some continuous function \(b\). Prove that if \(s\le t\) then \(x(t)\le x(s)\exp\int_s^t b(t)\,dt\).
Solution
Fix \(s\), and for \(u\ge s\) define \[ B(u)\coloneqq\int_s^u b(v)\,dv, \qquad y(u)\coloneqq x(u)e^{-B(u)}. \]
1. The function \(y\) is nonincreasing on \([s,\infty)\).
Since \(b\) is continuous, \[ B'(u)=b(u). \] Therefore \[ \begin{aligned} y'(u) &=e^{-B(u)} \left(x'(u)-b(u)x(u)\right). \end{aligned} \] The exponential factor is positive, while the hypothesis gives \[ x'(u)-b(u)x(u)\le0. \] Hence \(y'(u)\le0\), so \(y\) is nonincreasing.
2. If \(s\le t\), then\[ \boxed{ x(t) \le x(s)\exp\left(\int_s^t b(u)\,du\right) }. \]
By step 1, \[ y(t)\le y(s). \] Since \(B(s)=0\), this is \[ x(t)e^{-B(t)}\le x(s). \] Multiplying by the positive number \(e^{B(t)}\) gives \[ x(t)\le x(s)e^{B(t)} =x(s)\exp\left(\int_s^t b(u)\,du\right). \]
3. Q.E.D.
Step 2 is the required inequality.